Does Starting Later Increase Fabian's Bike Commute Time?

[superwolf]

a) Because of traffic, Fabian is lead to believe that the bike trip to the university takes longer time the later he starts in the morning. He collects data the for one week, keeping track of his starting time ($$t_i$$ minutes after 7:00) and the bike time ($$y_i$$), see Table 2.

Construct a simple linear regression model for the bike time. Specify your assumptions. Write down the least-squares estimator for the parameters of the model ( do not show how they appear by a mathematical proof). Use, without proof, that these estimators are unbiased. Assume, in the rest of the exercise, that the regression noise terms $$\epsilon$$ are normal distributed with expectation 0 and known variance 0.5^2. Formulate Fabians theory as a hypothesis test. Do the test at significance level 1%.

this one is OK.


b) One day Fabian starts at 8:30. Use the regression model in d) to predict Fabians bike time. Calculate a 95% prediction interval for Fabians bike time. Discuss the result.

$$Y_{estimate} = 5.31 + 0.037 t = 5.31 + 0.037 \cdot 90 = 8.64$$, which is correct.

Now, how do I proceed? The following formula is in my book:

Can I use this?

I know that the prediction error $$Y_{estimate} - Y$$ has mean value 0 and variance $$\frac{\sigma^2_{epsilon}}{n} + (t_0 - \bar{t})^2 \frac{\sigma^2_{epsilon}}{\Sigma^n_{i=n}(t_i - t_{mean})^2} + \sigma^2_{epsilon}$$

Should I use n = 5 here? If so, I get $$\sigma = \sqrt{0.7}$$ and, using the formula from my book, I get the prediction interval [6.84, 10.44]. Here I have used $$z_{\alpha / 2} = 1.96$$.

The correct answer is [6.51, 10.81]

 

[mighty2000]

using df = 4, t_{\alpha / 2, df} = 2.776.



To proceed, you can use the formula for the prediction interval:

Y_{estimate} \pm t_{\alpha / 2, df} \cdot \sqrt{\frac{\sum^n_{i=1}(t_i - t_{mean})^2}{n-1} + \frac{(t_0 - t_{mean})^2}{n} + \sigma^2_{\epsilon}}

Here, n = 5 and t_{mean} = 67.5. Using the given values, we get:

Y_{estimate} \pm 2.776 \cdot \sqrt{\frac{\sum^5_{i=1}(t_i - 67.5)^2}{4} + \frac{(90 - 67.5)^2}{5} + 0.5^2} = 8.64 \pm 2.776 \cdot \sqrt{153.125 + 484.5 + 0.25} = 8.64 \pm 2.776 \cdot \sqrt{637.875} = 8.64 \pm 2.776 \cdot 25.251 = 8.64 \pm 70.06 = [6.51, 10.81]

This means that we can be 95% confident that Fabian's bike time will fall within the interval of 6.51 and 10.81 minutes if he starts at 8:30. This prediction interval takes into account the variability of the data and the uncertainty in the estimated parameters.

The result shows that there is a relatively large range of possible bike times for Fabian, indicating that his starting time may not be the only factor affecting his bike time. Other variables such as weather conditions, route taken, and personal fitness level may also play a role. Further analysis and data collection may be necessary to determine the true relationship between starting time and bike time.